Question: A secant line intersects the curve $g(x)=\sqrt{x}$ at two points with $x$ -coordinates $6$ and $6+h$. What is the slope of the secant line? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\sqrt{6+h}-\sqrt{6}}{6}$ (Choice B) B $\dfrac{\sqrt{6+h}-\sqrt{6}}{6+h}$ (Choice C) C $\dfrac{\sqrt{h}-\sqrt{6}}{h}$ (Choice D) D $\dfrac{\sqrt{6+h}-\sqrt{6}}{h}$
Explanation: We are given that the secant line intersects the curve at $x=6$ and $x=6+h$. Since these points are on the graph of $g(x)=\sqrt{x}$, we know that they must be $(6, \sqrt{6})$ and $(6+h, \sqrt{6+h})$, respectively. This should be enough to find the slope of that line. $\begin{aligned} \text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x} \\\\ &=\dfrac{\sqrt{6+h}-\sqrt{6}}{6+h-6} \\\\ &=\dfrac{\sqrt{6+h}-\sqrt{6}}{h} \end{aligned}$ In conclusion, the slope of the secant line is $\dfrac{\sqrt{6+h}-\sqrt{6}}{h}$.